Total Submission(s): 547 Accepted Submission(s): 235
One day, MJF takes a stack of cards and talks to him: let's play a game and if you win, you can get all these cards. MJF randomly assigns these cards into n heaps, arranges in a row, and sets a value on each heap, which is called "penalty value".
Before the game starts, WYJ can move the foremost heap to the end any times.
After that, WYJ takes the heap of cards one by one, each time he needs to move all cards of the current heap to his hands and face them up, then he turns over some cards and the number of cards he turned is equal to the penaltyvalue.
If at one moment, the number of cards he holds which are face-up is less than the penaltyvalue, then the game ends. And WYJ can get all the cards in his hands (both face-up and face-down).
Your task is to help WYJ maximize the number of cards he can get in the end.So he needs to decide how many heaps that he should move to the end before the game starts. Can you help him find the answer?
MJF also guarantees that the sum of all "penalty value" is exactly equal to the number of all cards.
For each test case:
the first line is an integer n (1≤n≤106), denoting n heaps of cards;
next line contains n integers, the ith integer ai (0≤ai≤1000) denoting there are ai cards in ith heap;
then the third line also contains n integers, the ith integer bi (1≤bi≤1000) denoting the "penalty value" of ith heap is bi.
思路:根据题意,牌的数量和消耗的费用是相同的,所以总有一种顺序能够取到所有的牌:
我们首先从顺序头开始扫描求和,出现和为负的情况时做一个标记,然后和重置为0,继续向后扫描,最后把所有和为负的长度相加就是需要反转的次数
(来自队友的思路,听完以后豁然开朗,代码不到20行!)
#include<stdio.h>
int a[1000005],b[1000005],c[1000005];
int main(void){
int test;
while(scanf("%d",&test)!=EOF){
int i;
for(i=1;i<=test;i++)
scanf("%d",&a[i]);
for(i=1;i<=test;i++){
scanf("%d",&b[i]);
c[i]=a[i]-b[i];
}
long long sum=0,head=1;
for(i=1;i<=test;i++){
if(sum+c[i]<0){
head=i+1;
sum=0;
}
else sum+=c[i];
}
printf("%lld\n",head-1);
}
return 0;
}